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8 Geometric proofs
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Stage 5.1-3

8.06 Tangents and secants

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Lesson

Angles and tangents

A tangent to a circle touches the circle at precisely one point. We call this point the point of contact, or point of tangency.

A circle with centre O and radius OA with a tangent at A. The radius and tangent are perpendicular.

An important property of tangents is that they always form right angles with the radius drawn to the point of contact.

Tangents also make angles with chords, and the angles they make are equal to the angles the chords form in alternate segments:

Two circles with tangents and alternate segments highlighted. Ask your teacher for more information.

We will now prove that the angle between a tangent and a chord drawn to the point of contact is equal to the angle in the alternate segment.

The two tangents drawn to points A and B on a circle from an external point P. P A equals P B.

Two tangents that are drawn from the same point from outside of a circle will meet on opposite sides of the circle. The distances from the external point to each contact point will always be equal.

In the diagram, PA=PB, which means \triangle PAB is isosceles. Since AB is the base of this isosceles triangle, we can also conclude that \angle PAB=\angle PBA.

Examples

Example 1

In the diagram below, AC is a tangent to the circle with centre O:

A circle with centre O and tangent A C touching the circle at point B. The radius O B makes an angle x with the tangent.

Solve for x. Show all working and reasoning.

Worked Solution
Create a strategy

Use the property that tangents and radii form right angles at the point of contact on the circumference.

Apply the idea

The tangent AC and the radius OB are perpendicular. So:

\displaystyle x\displaystyle =\displaystyle 90(A tangent to a circle is perpendicular to the radius)
Idea summary
A circle with point O and radius OA which is perpendicular to the tangent at A.

A tangent to a circle is perpendicular to the radius drawn to the point of contact.

An angle between a tangent and and equal angle in the alternate segment of triangle A B C inscribed in a circle.

The angle between a tangent and a chord drawn to the point of contact is equal to the angle in the alternate segment.

The two tangents drawn to points A and B on a circle from an external point P. P A equals P B.

The two tangents drawn to a circle from an external point are equal in length.

Secants, chords and tangents

Two chords intersect each other inside a circle with line segments labelled as Y, B, A, and X.

If two chords intersect each other inside a circle, the products of their segment lengths are equal: a\times b=x\times y

A tangent A and secant with segments Y and X on a circle.

A secant is a line that intersects a circle at two points, unlike a tangent that only intersects a circle at one point. If a tangent and a secant are drawn to a circle from an exterior point, then the square of the distance from the exterior point to the point of contact along the tangent is equal to the product of the lengths of the exterior point to each intersection point along the secant: a^2 =x\times \left(x+y\right)

Two secants in a circle with segments labelled as B, A, Y, and X.

Similarly, if two secants are drawn from the same exterior point, the product of the distances from the exterior point to each intersection along each secant will be equal:a\times \left(a+b\right)=x\times \left(x+y\right)

Examples

Example 2

In the given diagram, CD is a tangent to the circle:

CD is a tangent to the circle, A B C is a secant to the circle, and triangle D A B is inscribed in a circle.

Prove that CD^{2} = CA \times CB using similar triangles.

Worked Solution
Create a strategy

Prove that \triangle CDA and \triangle CBD are similar.

Apply the idea
To prove: CD^{2} = CA \times CB
StatementsReasons
1.\angle CAD =\angle CDB(Angles in alternate segments)
2.\angle ACD is common\text{}
3.\triangle CDA ||| \triangle CBD(Equiangular)
4.\dfrac{CD}{CA}=\dfrac{CB}{CD} (Corresponding sides in similar triangles)
5.CD^2=CA\times CB\text{}

Example 3

In the diagram, chords AB and DC are produced to an external point P at which they intersect:

A circle with 2 chords A B and D C produced to an external point P. A B C D is a cyclic quadrilateral.

Prove that PB \times PA = PC \times PD using similar triangles.

Worked Solution
Create a strategy

Prove that \triangle PBC and \triangle PAD are similar.

Apply the idea
To prove: PB \times PA = PC \times PD
StatementsReasons
1.\angle APD is common\text{}
2.\angle BCP=180-\angle BCD(Angles on a straight line)
3.\angle DAP=180-\angle BCD (Opposite angles in a cyclic quadrilateral)
4.\angle BCP=\angle DAP\text{}
5.\triangle PBC ||| \triangle PAD(Equiangular)
6.\dfrac{PA}{PD}=\dfrac{PC}{PB} (Corresponding sides in similar triangles)
7.PB\times PA=PC\times PD\text{}

Example 4

In the diagram, points A, B, C and D lie on a circle, and AC and BD intersect at point E:

Points A, B, C, D lie on a circle in which chords A C and B D intersect at point E. Chords A D and B C are also drawn.

Prove that AE \times EC = BE \times ED using similar triangles.

Worked Solution
Create a strategy

Prove that \triangle AED and \triangle BEC are similar.

Apply the idea
To prove: AE \times EC = BE \times ED
StatementsReasons
1.\angle AED=\angle BEC(Vertically opposite angles)
2.\angle ADE=\angle BCE(Angles in the same segment)
3.\triangle AED ||| \triangle BEC (Equiangular)
4.\dfrac{AE}{BE}=\dfrac{ED}{EC}(Corresponding sides in similar triangles)
5.AE\times EC=BE\times ED\text{}
Idea summary
Three set of theorems about circles, length of secants, chords and tangents. Ask your teacher for more information.

Outcomes

MA5.3-17MG

applies deductive reasoning to prove circle theorems and to solve related problems

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